Today, the 14th of January 2014, we had a special session of our Theory Lunch. I spoke about ultrafilters and how they allow to generalize the notion of limit.

Consider the space of bounded sequences of real numbers, together with the supremum norm. We would like to define a notion of limit which holds for *every* and satisfies the well known properties of standard limit:

*Linearity:*.*Homogeneity:*.*Monotonicity:*if for every then .*Nontriviality:*if for every then .*Consistency:*if the limit exists in the classical sense, then the two notions coincide.

The consistency condition is reasonable also because it avoids trivial cases: if we fix and we define the limit of the sequence as the value , then the first four properties are satisfied.

Let us recall the classical definition of limit: we say that converges to if and only if, for every , the set of values such that is *cofinite*, *i.e.*, has a finite complement: the inequality can be satisfied at most for finitely many values of . The family of cofinite subsets of (in fact, of any set ) has the following properties:

*Upper closure:*if and then .*Meet stability:*if then .

A family of subsets of with the two properties above is called a *filter* on . An immediate example is the *trivial filter* ; another example is the *improper filter* . The family of cofinite subset of is called the *Fréchet filter* on . The Fréchet filter is not the improper one if and only if is infinite.

An *ultrafilter* on is a filter on satisfying the following additional conditions:

*Properness:*.*Maximality:*for every , either or .

For example, if , then is an ultrafilter on , called the *principal ultrafilter* generated by . Observe that : if we say that is *free*. These are, in fact, the only two options.

**Lemma 1.** For a proper filter to be an ultrafilter, it is necessary and sufficient that it satisfies the following condition: for every and nonempty , if then for at least one .

*Proof:* It is sufficient to prove the thesis with . If with , then is a proper filter that properly contains . If the condition is satisfied, for every which is neither nor we have , thus either or .

**Theorem 1.** Every nonprincipal ultrafilter is free. In addition, an ultrafilter is free if and only if it extends the Fréchet filter. In particular, every ultrafilter over a finite set is principal.

*Proof:* Let be a nonprincipal ultrafilter. Let : then , so either there exists such that and , or there exists such that and . In the first case, ; in the second case, we consider and reduce to the first case. As is arbitrary, is free.

Now, for every the set belongs to but not to : therefore, no principal ultrafilter extends the Fréchet filter. On the other hand, if is an ultrafilter, is finite, and , then by maximality, hence for some because of Lemma 1, thus cannot be a free ultrafilter.

So it seems that free ultrafilters are the right thing to consider when trying to expand the concept of limit. There is an issue, though: we have not seen any single example of a free ultrafilter; in fact, we do not even (yet) know whether free ultrafilters do exist! The answer to this problem comes, in a shamelessly nonconstructive way, from the following

**Ultrafilter lemma.** Every proper filter can be extended to an ultrafilter.

The ultrafilter lemma, together with Theorem 1, implies the existence of free ultrafilters on every infinite set, and in particular on . On the other hand, to prove the ultrafilter lemma the Axiom of Choice is required, in the form of Zorn’s lemma. Before giving such proof, we recall that a family of sets has the *finite intersection property* if every finite subfamily has a nonempty intersection: every proper filter has the finite intersection property.

*Proof of the ultrafilter lemma.* Let be a proper filter on and let be the family of the collections of subsets of that extend and have the finite intersection property, ordered by inclusion. Let be a totally ordered subfamily of : then extends and has the finite intersection property, because for every finitely many there exists by construction such that .

By Zorn’s lemma, has a maximal element , which surely satisfies and . If and , then still has the finite intersection property, therefore by maximality. If then still has the finite intersection property, therefore again by maximality.

Suppose, for the sake of contradiction, that there exists such that and : then neither nor have the finite intersection property, hence there exist such that . But means , and means : therefore,

against having the finite intersection property.

We are now ready to expand the idea of limit. Let be a metric space and let be an ultrafilter on : we say that is the *ultralimit* of the sequence along if for every the set

belongs to . (Observe how, in the standard definition of limit, the above set is required to belong to the Fréchet filter.) If this is the case, we write

Ultralimits, if they exist, are unique and satisfy our first four conditions. Moreover, the choice of a principal ultrafilter corresponds to the trivial definition . So, what about free ultrafilters?

**Theorem 2.** Every bounded sequence of real numbers has an ultralimit along every free ultrafilter on .

*Proof:* It is not restrictive to suppose for every . Let be an arbitrary, but fixed, free ultrafilter on . We will construct a sequence of closed intervals , , such that and for every . By the Cantor intersection theorem it will be : we will then show that .

Let . Let be either or , chosen according to the following criterion: . If both halves satisfy the criterion, then we just choose one once and for all. We iterate the procedure by always choosing as one of the two halves of such that .

Let . Let , and let be so large that : then , thus . As the smaller set belongs to , so does the larger one.

We have thus almost achieved our original target: a notion of limit which applies to every bounded sequence of real numbers. Such notion will depend on the specific free ultrafilter we choose: but it is already very reassuring that such a notion exists at all! To complete our job we need one more check: we have to be sure that the definition is consistent with the classical one. And this is indeed what happens!

**Theorem 3.** Let be a sequence of real numbers and let . Then in the classical sense if and only if for every free ultrafilter on .

To prove Theorem 3 we make use of an auxiliary result, which is of interest by itself.

**Lemma 2.** Let be the family of collections of subsets of that have the finite intersection property. The maximal elements of are precisely the ultrafilters.

*Proof:* Every ultrafilter is clearly maximal in . If is maximal in , then it is clearly proper and upper closed, and we can reason as in the proof of the ultrafilter lemma to show that it is actually an ultrafilter.

*Proof of Theorem 3:* Suppose does not converge to in the classical sense. Fix such that the set is infinite. Then the family has the finite intersection property: an ultrafilter latex \mathcal{V}$ must be free. Then , and does not have an ultralimit along .

The converse implication follows from the classical definition of limit, together with the very notion of free ultrafilter.

Theorem 3 does hold for sequences of real numbers, but does not extend to arbitrary metric spaces. In fact, the following holds, which we state without proving.

**Theorem 4.** Let be a metric space. The following are equivalent.

- For some free ultrafilter on , every sequence in has an ultralimit along .
- For every free ultrafilter on , every sequence in has an ultralimit along .
- is compact.

Ultrafilters are useful in many other contexts. For instance, they are used to construct *hyperreal numbers*, which in turn allow a rigorous definition of infinitesimals and the foundation of calculus over those. But this might be the topic for another Theory Lunch talk.

I fixed the text a little bit. I also corrected a crucial error in the fifth paragraph, which reversed the role of the words “finite” and “cofinite”.